{"id":5494,"date":"2022-06-02T18:51:58","date_gmt":"2022-06-02T18:51:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=5494"},"modified":"2022-11-01T05:18:35","modified_gmt":"2022-11-01T05:18:35","slug":"flux-form-of-greens-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/flux-form-of-greens-theorem\/","title":{"raw":"Flux Form of Green\u2019s Theorem","rendered":"Flux Form of Green\u2019s Theorem"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Apply <\/span>the flux form of Green\u2019s theorem.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n
    The circulation form of Green\u2019s theorem relates a double integral over region [latex]D[\/latex] to line integral [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}ds[\/latex], where [latex]C[\/latex] is the boundary of [latex]D[\/latex]. The flux form of Green\u2019s theorem relates a double integral over region [latex]D[\/latex] to the flux across boundary [latex]C[\/latex]. The flux of a fluid across a curve can be difficult to calculate using the flux line integral. This form of Green\u2019s theorem allows us to translate a difficult flux integral into a double integral that is often easier to calculate.<\/div>\r\n
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    theorem: Green's theorem, flux form<\/h3>\r\n

    Let [latex]D[\/latex] be an open, simply connected region with a boundary curve [latex]C[\/latex] that is a piecewise smooth, simple closed curve that is oriented counterclockwise (Figure 1). Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives on an open region containing [latex]D[\/latex]. Then,<\/p>\r\n

    [latex]\\displaystyle\\oint{\\bf{F}}\\cdot{\\bf{N}}ds=\\displaystyle\\iint_DP_x+Q_ydA[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n[caption id=\"attachment_5288\" align=\"aligncenter\" width=\"493\"]\"<img Figure 1. The flux form of Green\u2019s theorem relates a double integral over region [latex]D[\/latex] to the flux across curve [latex]C[\/latex].[\/caption]\r\n

    Because this form of Green\u2019s theorem contains unit normal vector [latex]\\bf{N}[\/latex], it is sometimes referred to as the\u00a0normal form<\/em>\u00a0of Green\u2019s theorem<\/span>.<\/p>\r\n\r\n

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    Proof<\/h3>\r\n

    Recall that [latex]\\displaystyle\\oint{\\bf{F}}\\cdot{\\bf{N}}ds=\\displaystyle\\oint_C-Qdx+Pdy[\/latex]. Let [latex]M=-Q[\/latex] and [latex]N=P[\/latex]. By the circulation form of Green\u2019s theorem,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\oint_C-Qdx+Pdy&=\\displaystyle\\oint_CMdx+nDy \\\\\r\n&=\\displaystyle\\iint_DN_x-M_ydA \\\\\r\n&=\\displaystyle\\iint_DP_x-(-Q)_ydA \\\\\r\n&=\\displaystyle\\iint_DP_x+Q_ydA\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n

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    Example: Applying Green's theorem for flux across a circle<\/h3>\r\nLet [latex]C[\/latex] be a circle of radius [latex]r[\/latex] centered at the origin (Figure 2) and let [latex]{\\bf{F}}(x,y)=\\langle{x},y\\rangle[\/latex]. Calculate the flux across [latex]C[\/latex].\r\n\r\n[caption id=\"attachment_5290\" align=\"aligncenter\" width=\"717\"]\"<img Figure 2. Curve [latex]C[\/latex] is a circle of radius [latex]r[\/latex] centered at the origin.[\/caption][reveal-answer q=\"345083481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"345083481\"]\r\n

    Let [latex]D[\/latex] be the disk enclosed by [latex]C[\/latex]. The flux across [latex]C[\/latex] is [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{N}}ds[\/latex]. We could evaluate this integral using tools we have learned, but Green\u2019s theorem makes the calculation much more simple. Let [latex]P(x, y)=x[\/latex] and [latex]Q(x, y)=y[\/latex] so that [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex]. Note that [latex]P_x=1=Q_y[\/latex], and therefore [latex]P_x+Q_y=2[\/latex]. By Green\u2019s theorem,<\/p>\r\n

    [latex]\\displaystyle\\int_C{\\bf{F}}\\bullet{\\bf{N}}=\\displaystyle\\int \\ \\displaystyle\\int_D2dA=2\\displaystyle\\int \\ \\displaystyle\\int_DdA[\/latex].<\/p>\r\n

    Since [latex]\\displaystyle\\int_DdA[\/latex] is the area of the circle, [latex]\\displaystyle\\int_DdA=\\pi{r}^2[\/latex]. Therefore, the flux across [latex]C[\/latex] is\u00a0[latex]2\\pi{r}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    Example: applying green's theorem for flux across a triangle<\/h3>\r\nLet [latex]S[\/latex] be the triangle with vertices [latex](0, 0)[\/latex], [latex](1, 0)[\/latex], and [latex](0, 3)[\/latex] oriented clockwise (Figure 3). Calculate the flux of [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{x}^2+e^y,x+y\\rangle[\/latex] across [latex]S[\/latex].\r\n\r\n[caption id=\"attachment_5295\" align=\"aligncenter\" width=\"679\"]\"<img Figure 3. Curve [latex]S[\/latex] is a triangle with vertices [latex](0, 0)[\/latex], [latex](1, 0)[\/latex], and [latex](0, 3)[\/latex] oriented clockwise.[\/caption][reveal-answer q=\"398702952\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"398702952\"]\r\n

    To calculate the flux without Green\u2019s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green\u2019s theorem to translate the flux line integral into a single double integral is much more simple.<\/p>\r\n

    Let [latex]D[\/latex] be the region enclosed by [latex]S[\/latex]. Note that [latex]P_x=2x[\/latex] and [latex]Q_y=1[\/latex] therefore, [latex]P_x+Q_y=2x+1[\/latex]. Green\u2019s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{N}}ds=-\\displaystyle\\oint_{-S}{\\bf{F}}\\cdot{\\bf{N}}ds[\/latex] and [latex]-S[\/latex] is oriented counterclockwise. By Green\u2019s theorem, the flux is<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{N}}ds&=\\displaystyle\\oint_{-S}{\\bf{F}}\\cdot{\\bf{N}}ds \\\\\r\n&=-\\displaystyle\\iint_D(P_x+Q_y)dA \\\\\r\n&=-\\displaystyle\\iint_D(2x+1)dA\r\n\\end{aligned}[\/latex]<\/p>\r\n

    Notice that the top edge of the triangle is the line [latex]y=-3x=3[\/latex] Therefore, in the iterated double integral, the\u00a0y<\/em>-values run from [latex]y=0[\/latex] to [latex]y=-3x+3[\/latex], and we have<\/p>\r\n

    [latex]\\begin{aligned}\r\n-\\displaystyle\\iint_D(2x+1)dA&=-\\displaystyle\\int_0^1\\displaystyle\\int_0^{-3x+3}(2x+1)dydx \\\\\r\n&=--\\displaystyle\\int_0^1(2x+1)(-3x+3)dx=-\\displaystyle\\int_0^1(-6x^2+3x+3)dx \\\\\r\n&=-\\left[-2x^3+\\frac{3x^2}2+3x\\right]_0^1=-\\frac52\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nCalculate the flux of [latex]{\\bf{F}}(x,y)=\\langle{x}^3,y^3\\rangle[\/latex]\u00a0 across a unit circle oriented counterclockwise.\r\n\r\n[reveal-answer q=\"457134886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457134886\"]\r\n\r\n[latex]\\frac{3\\pi}2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n